Question

A stress tensor $\sigma$, with elements in MPa, is as given. The maximum value of the principal stress in MPa is \[ \sigma = \begin{bmatrix} 1 & 0 & \sqrt{2}
0 & 1 & 0
\sqrt{2} & 0 & 0 \end{bmatrix} \]

• A. 2.0
• B. $\sqrt{2}$
• C. 1.0
• D. 0.0

Hint:

Principal stresses are found by solving the eigenvalue problem \(\det(\sigma - \lambda I) = 0\). For symmetric stress tensors, eigenvalues are always real. The maximum eigenvalue corresponds to the maximum principal stress.

Answer 1

The Correct Option is A

Step 1: Principal stresses
Principal stresses are the eigenvalues of the stress tensor $\sigma$. We solve: \[ \det(\sigma - \lambda I) = 0 \] Step 2: Write matrix $(\sigma - \lambda I)$
\[ \sigma - \lambda I = \begin{bmatrix} 1-\lambda & 0 & \sqrt{2}
0 & 1-\lambda & 0
\sqrt{2} & 0 & -\lambda \end{bmatrix} \] Step 3: Determinant expansion
\[ \det(\sigma - \lambda I) = (1-\lambda)\big[ (1-\lambda)(-\lambda) - (0)(0) \big] - 0 + \sqrt{2}\big[ 0 - (1-\lambda)\sqrt{2} \big] \] Simplify: \[ = (1-\lambda)(-(\lambda)(1-\lambda)) + \sqrt{2} \big[ - (1-\lambda)\sqrt{2} \big] \] \[ = -(1-\lambda)^2 \lambda - 2(1-\lambda) \] Factor: \[ = -(1-\lambda)\big[ (1-\lambda)\lambda + 2 \big] \] Step 4: Characteristic equation
\[ -(1-\lambda)(\lambda^2 - \lambda + 2) = 0 \] So eigenvalues: \[ \lambda_1 = 1, \quad \lambda_{2,3} = \frac{1 \pm \sqrt{1-8}}{2} = \frac{1 \pm i\sqrt{7}}{2} \] Step 5: Reality check
Stress tensors are symmetric $\Rightarrow$ eigenvalues must be real. Let us re-check carefully. \[ \sigma = \begin{bmatrix} 1 & 0 & \sqrt{2}
0 & 1 & 0
\sqrt{2} & 0 & 0 \end{bmatrix} \] is symmetric. Let's compute again. Step 6: Determinant carefully
\[ \det(\sigma - \lambda I) = \begin{vmatrix} 1-\lambda & 0 & \sqrt{2}
0 & 1-\lambda & 0
\sqrt{2} & 0 & -\lambda \end{vmatrix} \] Expanding along the second row: \[ = (1-\lambda)\begin{vmatrix} 1-\lambda & \sqrt{2}
\sqrt{2} & -\lambda \end{vmatrix} \] \[ = (1-\lambda)[(1-\lambda)(- \lambda) - ( \sqrt{2}\cdot \sqrt{2})] \] \[ = (1-\lambda)[ -\lambda(1-\lambda) - 2 ] \] \[ = (1-\lambda)( -\lambda + \lambda^2 - 2 ) \] \[ = (1-\lambda)(\lambda^2 - \lambda - 2) \] \[ = (1-\lambda)(\lambda - 2)(\lambda + 1) \] Step 7: Eigenvalues
Thus: \[ \lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = -1 \] Step 8: Maximum principal stress
The maximum principal stress = largest eigenvalue = \[ \boxed{2.0 \, \text{MPa}} \]